\subsection{The M\"obius group}
M\"obius groups are an analogous concept to permutation groups, but on the infinite set of the complex numbers.
A M\"obius transformation \(f\) is defined as follows:
\[
	f: \Chat \to \Chat;\quad f(z) = \frac{az + b}{cz + d};\quad a, b, c, d \in \mathbb C;\quad ad-bc \neq 0
\]
The reason for the restriction that \(ad-bc\neq 0\) is that \(ad=bc\) implies that \(f\) is a constant value for all \(z\).
Note that \(\Chat\) is known as the `extended complex plane', defined as the complex plane together with a point at infinity, denoted \(\infty\).
There are some special points related to M\"obius transformations:
\begin{itemize}
	\item \(f\qty(\frac{-d}{c})\) is defined to be \(\infty\).
	      This is because the denominator of the fraction would be zero.
	\item \(f(\infty)\) is defined to be \(\frac{a}{c}\) if \(c \neq 0\).
	      This is because as the length of \(z\) increases to infinity, the constant terms \(b\) and \(d\) vanish.
	      However, if \(c = 0\), then the numerator explodes to infinity as the denominator remains constant, so \(f(\infty) = \infty\) in this case.
\end{itemize}

\begin{lemma}
	M\"obius transformations are bijections from \(\Chat \to \Chat\).
\end{lemma}
\begin{proof}
	We can prove this by evaluating \(f(f^{-1}(z))\) and \(f^{-1}(f(z))\) at all \(z\), taking into account all the special points.
	The entire proof is not written here, but it suffices to substitute every special point and a generic \(z\) into both of these expressions, and show that they equal \(z\) in all cases.
\end{proof}

\begin{theorem}
	The set \(\mathcal M\) of M\"obius maps forms a group under composition of functions.
\end{theorem}
\begin{proof}
	We must check each of the group axioms, and we begin with closure.
	Let \(f_1(z) = \frac{a_1 z + b_1}{c_1 z + d_1}; f_2(z) = \frac{a_2 z + b_2}{c_2 z + d_2}\).
	To compose these functions, we first ignore the special points and then check them individually later.
	\begin{align*}
		(f_2 \circ f_1)(z) & = f_2(f_1(z))                                                                                                               \\
		                   & = \frac{a_2 \left( \frac{a_1 z + b_1}{c_1 z + d_1} \right) + b_2}{c_2 \left( \frac{a_1 z + b_1}{c_1 z + d_1} \right) + d_2} \\
		                   & = \frac{(a_1 a_2 + b_2 c_1)z + (a_2 b_1 + b_2 d_1)}{(c_2 a_1 + d_2 c_1)z + (c_2 b_1 + d_1 d_2)}                             \\
		                   & =: \frac{az + b}{cz + d}
	\end{align*}
	Note that \(ad-bc = (a_1 a_2 + b_2 c_1)(c_2 b_1 + d_1 d_2) - (a_2 b_1 + b_2 d_1)(c_2 a_1 + d_2 c_1) = (a_1 d_1 - b_1 c_1)(a_2 d_2 - b_2 c_2)\) which is the product of two nonzero numbers, which is therefore nonzero.
	Now we will check all the special points.
	\begin{align*}
		(f_2 \circ f_1)(\infty)                                         & = f_2\left(\frac{a_1}{c_1}\right)                                                           \\
		                                                                & = \frac{a_2 \left( \frac{a_1}{c_1} \right) + b_2}{c_2 \left( \frac{a_1}{c_1} \right) + d_2} \\
		                                                                & = \frac{a_1 a_2 + b_2 c_1}{c_2 a_1 + d_2 c_1}                                               \\
		                                                                & = \frac{a}{c}                                                                               \\
		(f_2 \circ f_1)(\infty)                                         & = f_2(\infty) = \frac{a_2}{c_2}                                                             \\
		(f_2 \circ f_1)\left(f^{-1}\left(\frac{-d_2}{c_2}\right)\right) & = f_2\left( \frac{-d_2}{c_2} \right)                                                        \\
		                                                                & = \infty
	\end{align*}
	Note that each of these results matches up with our intuitive understanding of infinity in the limit, for instance \((f_2 \circ f_1)(\infty) = \frac{a}{c}\), where na\"\i{}vely we might assume \((f_2 \circ f_1)(\infty) = \frac{a \cdot \infty + b}{c \cdot \infty + d} = \frac{a}{c}\).

	Now we may prove the other group axioms hold for \(\mathcal M\).
	Clearly there is an identity element \(f(z) = \frac{1z + 0}{0z + 1}\).
	We know that there are always inverses because \(f\) is a bijection.
	Finally, we know that all M\"obius maps obey the associative law because function composition is always associative.
	So \(\mathcal M\) is a group.
\end{proof}

\subsection{Properties of the M\"obius group}
When we are working with M\"obius groups, we use the following conventions:
\[
	\frac{1}{\infty} = 0;\quad \frac{1}{0} = \infty;\quad \frac{a\cdot\infty}{c\cdot\infty} = \frac{a}{c}
\]

Firstly, \(\mathcal M\) is not abelian.
For example, let \(f_1(z) = z + 1; f_2(z) = 2z\).
Then \((f_2 \circ f_1)(z) = 2z + 2\) and \((f_1 \circ f_2)(z) = 2z + 1\).

\begin{proposition}
	Every M\"obius transformation can be written as a composition of maps of the following forms:
	\begin{enumerate}
		\item \(f(z) = az\) where \(a\neq 0\).
		      This is a dilation or rotation.
		\item \(f(z) = z + b\).
		      This is a translation by \(b\).
		\item \(f(z) = \frac{1}{z}\).
		      This is an inversion.
	\end{enumerate}
\end{proposition}
\begin{proof}
	Let \(f(z) = \frac{az + b}{cz + d}\).
	Then if \(c \neq 0\), \(f(z)\) is given by
	\[
		z \xrightarrow{\text{(ii)}} z + \frac{d}{c} \xrightarrow{\text{(iii)}} \frac{1}{z + \frac{d}{c}} \xrightarrow{\text{(i)}} \frac{(-ad+bc)c^{-2}}{z + \frac{d}{c}} \xrightarrow{\text{(ii)}} \frac{a}{c} + \frac{(-ad+bc)c^{-2}}{z + \frac{d}{c}} = \frac{az + b}{cz + d}
	\]
	If \(c = 0\), \(f(z)\) is given by
	\[
		z \xrightarrow{\text{(i)}} \frac{a}{d}z \xrightarrow{\text{(ii)}} \frac{a}{d}z + \frac{b}{d} = \frac{az + b}{d}
	\]
\end{proof}
Note therefore that the set \(\mathcal S\) of all dilations, rotations, translations and inversions generates \(\mathcal M\), or in symbolic form, \(\genset {\mathcal S} = \mathcal M\).
